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HW12-solutions

# HW12-solutions - munoz(gm7794 HW12 radin(54915 This...

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munoz (gm7794) – HW12 – radin – (54915) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite the series 4 parenleftbigg 1 4 parenrightbigg 2 sin 3 5 - 4 parenleftbigg 1 4 parenrightbigg 3 sin 4 6 + 4 parenleftbigg 1 4 parenrightbigg 4 sin 5 7 + . . . using summation notation. 1. sum = summationdisplay k =3 parenleftbigg - 1 4 parenrightbigg k 1 4 sin k k + 2 correct 2. sum = 25 summationdisplay k =3 parenleftbigg - 1 4 parenrightbigg k 1 4 sin k k + 2 3. sum = summationdisplay k =1 parenleftbigg - 1 4 parenrightbigg k 4 sin( k + 2) k + 4 4. sum = 80 summationdisplay k =2 parenleftbigg 1 4 parenrightbigg k 4 sin( k + 1) k + 3 5. sum = summationdisplay k =3 parenleftbigg - 1 4 parenrightbigg k 1 4 sin k k + 1 Explanation: The given series is an infinite series, so two of the answers must be incorrect because they are finite series written in summation notation. Starting summation at k = 3 we see that the general term of the infinite series is a k = 4 parenleftbigg - 1 4 parenrightbigg k 1 sin k k + 2 . Consequently, sum = summationdisplay k =3 parenleftbigg - 1 4 parenrightbigg k 1 4 sin k k + 2 . 002 10.0 points Determine whether the series summationdisplay n =0 2 (cos ) parenleftbigg 2 3 parenrightbigg n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum - 5 6 2. convergent with sum - 6 5 3. convergent with sum 6 5 correct 4. convergent with sum 6 5. convergent with sum - 6 6. divergent Explanation: Since cos = ( - 1) n , the given series can be rewritten as an infinite geometric series summationdisplay n =0 2 parenleftbigg - 2 3 parenrightbigg n = summationdisplay n =0 a r n in which a = 2 , r = - 2 3 . But the series n =0 ar n is (i) convergent with sum a 1 - r when | r | < 1, and (ii) divergent when | r | ≥ 1. Consequently, the given series is convergent with sum 6 5 .

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munoz (gm7794) – HW12 – radin – (54915) 2 003 10.0 points Determine if the series summationdisplay k =1 5 + 2 k 3 k converges or diverges, and if it converges, find its sum. 1. converges with sum = 9 2 correct 2. converges with sum = 3 3. converges with sum = 5 4. converges with sum = 7 2 5. series diverges 6. converges with sum = 4 Explanation: An infinite geometric series n =1 a r n 1 (i) converges when | r | < 1 and has sum = a 1 - r , while it (ii) diverges when | r | ≥ 1 . Now summationdisplay k =1 5 3 k = summationdisplay k =1 5 3 parenleftBig 1 3 parenrightBig k 1 is a geometric series with a = r = 1 3 < 1. Thus it converges with sum = 5 2 , while summationdisplay k =1 2 k 3 k = summationdisplay k =1 2 3 parenleftBig 2 3 parenrightBig k 1 is a geometric series with a = r = 2 3 < 1. Thus it too converges, and it has sum = 2 . Consequently, being the sum of two conver- gent series, the given series converges with sum = 5 2 + 2 = 9 2 .
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HW12-solutions - munoz(gm7794 HW12 radin(54915 This...

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